What is 1/((1+2-1))*ln(1/(1-2)) ?

The solution to 1/((1+2-1))*ln(1/(1-2)) is -(ln(-1))/2

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simplify 1/((1+2-1))*ln(1/(1-2))

simplify

\frac{1}{( 1 + 2 - 1 )} \cdot ln (\frac{1}{1 - 2})

- \frac{ln ( - 1 )}{2}

Solution steps

\frac{1}{( 1 + 2 - 1 )} ln (\frac{1}{1 - 2})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{1}{1 - 2}) = ln (1) - ln (1 - 2)

= \frac{1}{1 + 2 - 1} (ln (1) - ln (1 - 2))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( ln ( 1 ) - ln ( 1 - 2 ) )}{1 + 2 - 1}

1 \cdot (ln (1) - ln (1 - 2)) = ln (1) - ln (- 1)

= \frac{ln ( 1 ) - ln ( - 1 )}{1 + 2 - 1}

Add/Subtract the numbers:

1 + 2 - 1 = 2

= \frac{ln ( 1 ) - ln ( - 1 )}{2}

Apply log rule:

lo g_{a} (1) = 0

= \frac{0 - ln ( - 1 )}{2}

0 - ln (- 1) = - ln (- 1)

= \frac{- ln ( - 1 )}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{ln ( - 1 )}{2}

\frac{1}{3} ln (\frac{540}{180})

ln (\frac{20}{21}) \cdot 70

\frac{1}{- 0.10986} ln (\frac{40}{60})

\frac{1}{2} lo g_{4} (x) + \frac{1}{2} lo g_{4} (y)

lo g_{4} (7 e)

What is 1/((1+2-1))*ln(1/(1-2)) ?
The solution to 1/((1+2-1))*ln(1/(1-2)) is -(ln(-1))/2